Affine Reflections, Coroot Translations, and Alcoves

Substituting the reflection formula from [F1] gives $t_{k{\alpha }^{\vee }}(s_{\alpha }(x))=x-B(x,\alpha ){\alpha }^{\vee }+k{\alpha }^{\vee }=x-(B(x,\alpha )-k){\alpha }^{\vee }=r_{\alpha ,k}(x)$, proving the translation form.

If $K=\mathrm{\varnothing }$ there are no walls meeting it. Otherwise fix $\alpha \in \mathrm{\Phi }$ and for each real $c>0$ put $U_c=\{x\in E:\mathrm{\mid }B(x,\alpha )\mathrm{\mid }<c\}$. For $x\in U_c$, Cauchy–Schwarz from [F6] shows that every $y$ with $d_B(x,y)<(c-\mathrm{\mid }B(x,\alpha )\mathrm{\mid })\mathrm{/}\mathrm{\parallel }\alpha {\mathrm{\parallel }}_B$ also lies in $U_c$, so $U_c$ is open. Its traces $U_c\cap K$ form an open cover of the subspace $K$, since $x\in U_{\mathrm{\mid }B(x,\alpha )\mathrm{\mid }+1}$; compactness gives finitely many parameters $c_0,\dots ,c_n$ with $K\subseteq {\bigcup }_i{U}_{c_i}$. Set $C_{\alpha }={\max }_i{c}_i$, which exists by [F4]; then $\mathrm{\mid }B(x,\alpha )\mathrm{\mid }<C_{\alpha }$ for all $x\in K$. By Archimedeanness choose a natural $N>C_{\alpha }$. If $H_{\alpha ,k}$ meets $K$, then $\mathrm{\mid }k\mathrm{\mid }<N$; [F7] identifies $\mathrm{\mid }k\mathrm{\mid }$ with a natural $j<N$, so the possible integers are on the finite list $0,\pm 1,\dots ,\pm (N-1)$. Thus only finitely many $k$ occur for this root, and finiteness of $\mathrm{\Phi }$ gives finitely many walls in total.

If $\mathrm{\Phi }=\mathrm{\varnothing }$, the arrangement is empty. Otherwise, for any $p\in E$ the finite intersection $U={\bigcap }_{\alpha \in \mathrm{\Phi }}B(p,1\mathrm{/}(2\mathrm{\parallel }\alpha {\mathrm{\parallel }}_B))$ is an open neighbourhood of $p$ by [F10]. By Cauchy–Schwarz, any wall $H_{\alpha ,k}$ meeting $U$ has $\mathrm{\mid }k-B(p,\alpha )\mathrm{\mid }<1\mathrm{/}2$; by [F7] at most one integer $k$ occurs for each root, so $U$ meets only finitely many walls. Each wall is closed: if $p\mathrm{\notin }{H}_{\alpha ,k}$ then the ball of radius $\mathrm{\mid }B(p,\alpha )-k\mathrm{\mid }\mathrm{/}(2\mathrm{\parallel }\alpha {\mathrm{\parallel }}_B)$ about $p$ misses it by the same inequality. Since only finitely many walls meet $U$, their union is closed; around any point outside the full union, remove this finite closed union from $U$ to obtain a neighbourhood avoiding every wall. Hence the full union is closed and its complement is open.

The formula $\psi (\lambda ,w)(x)=\lambda +w(x)$ and the multiplication in [F5] give $\psi (\lambda ,w)\circ \psi (\mu ,v)(x)=\lambda +w\mu +wv(x)=\psi (\lambda +w\mu ,wv)(x)$, so $\psi$ is a group homomorphism. If $\psi (\lambda ,w)={\mathrm{i}\mathrm{d}}_E$, evaluating at $0$ gives $\lambda =0$, after which $w(x)=x$ for all $x$ and $w=1$; thus $\psi$ is injective.

The coroot set ${\mathrm{\Phi }}^{\vee }$ is a reduced crystallographic root system: it is finite, nonzero and spanning because each coroot is a nonzero multiple of its root; reducedness follows from [F1] and the involution $({\alpha }^{\vee }{)}^{\vee }=\alpha$. Its root reflection is $s_{{\alpha }^{\vee }}=s_{\alpha }$, and orthogonality gives $s_{\alpha }({\beta }^{\vee })=(s_{\alpha }\beta {)}^{\vee }$, so it preserves ${\mathrm{\Phi }}^{\vee }$. Its Cartan numbers are $B(({\alpha }^{\vee }{)}^{\vee },{\beta }^{\vee })=B(\alpha ,{\beta }^{\vee })\in \mathbb{Z}$ by [F1]. Use the same regular vector that defines the given positive roots; coroots have the same signs as their roots. If $\beta ={\sum }_s{n}_s{\alpha }_s$ is positive, then ${\beta }^{\vee }={\sum }_s{n}_{s}B({\alpha }_s,{\alpha }_s)\mathrm{/}B(\beta ,\beta ){\alpha }_s^{\vee }$ has nonnegative real coordinates. Thus an equality ${\alpha }_s^{\vee }={\beta }^{\vee }+{\gamma }^{\vee }$ for positive coroots forces both $\beta ,\gamma$ onto the line of ${\alpha }_s$ by coordinatewise nonnegativity. Reducedness then forces both to equal ${\alpha }_s$, contradicting the equality. Hence every ${\alpha }_s^{\vee }$ is dual-simple. In applying [F3] to this dual system, the sign split used in its linear-independence argument is justified as follows: for any finite family of positive roots ${\beta }_j$, if ${\sum }_j{c}_j{\beta }_j^{\vee }=0$ with $c_j\ge 0$ and some $c_j>0$, then $B\text{\hspace{0.17em}⁣}(v,{\sum }_j{c}_j{\beta }_j^{\vee })={\sum }_j{c}_j\frac{2B(v,{\beta }_j)}{B({\beta }_j,{\beta }_j)}>0,$ contradiction; negating rules out a nonzero relation with all $c_s\le 0$. Thus every nontrivial relation among the dual simple roots has both positive and negative coefficients, even before identifying the complete dual simple set; this is the case treated in the remainder of that signed-integral basis proof. Apply the signed integral basis theorem in [F3] to the now-verified dual root system: its simple set is a basis with exactly $\dim E=\mathrm{\mid }S\mathrm{\mid }$ elements, so it equals $\{{\alpha }_s^{\vee }:s\in S\}$. Every coroot is therefore an integral combination of the simple coroots, proving $Q^{\vee }={⨁}_s\mathbb{Z}{\alpha }_s^{\vee }$. This includes the empty system and all orthogonal components.

By [F1], $B(r_{\alpha ,k}(x),\alpha )=B(x,\alpha )-(B(x,\alpha )-k)B({\alpha }^{\vee },\alpha )=2k-B(x,\alpha )$. Thus $B(r_{\alpha ,k}(x),\alpha )-k=k-B(x,\alpha )$, and substituting this into the formula for a second application gives $r_{\alpha ,k}^2(x)=x$. Also $r_{\alpha ,k}(x)=x$ exactly when $(B(x,\alpha )-k){\alpha }^{\vee }=0$, which by ${\alpha }^{\vee }\mathrm{\ne }0$ is equivalent to $x\in H_{\alpha ,k}$.

Let $C$ be an alcove and $x\in C$. Since the complement of the wall union is open by step 1.3, a ball around $x$ lies in it; the ball is path-connected by straight segments and hence connected by [F11], so it lies in the connected component $C$. Thus $C$ is open. For any wall $H_{\beta ,l}$, both strict halfspaces are open by the Cauchy–Schwarz estimate in step 1.2 and [F10], so $C\cap \{y:B(y,\beta )<l\}$ and $C\cap \{y:B(y,\beta )>l\}$ are relatively open and partition $C$; connectedness forces one to be empty. Therefore any $x,y\in C$ lie on the same strict side of every wall, and linearity of $B(-,\beta )$ shows their segment avoids every wall. That segment is connected, meets $C$, and lies in the complement, so it lies in $C$; hence $C$ is convex.

If $\mathrm{\Phi }=\mathrm{\varnothing }$, then $E=\{0\}$ by [F12]; the arrangement is empty and $E$ itself is the unique alcove, so the local-finiteness claim is immediate. Otherwise fix $p\in E$ and take the neighbourhood $U$ from step 1.3. It is convex: if $x,y\in U$, $t\in [0,1]$, and $\alpha \in \mathrm{\Phi }$, then membership in $B(p,1\mathrm{/}(2\mathrm{\parallel }\alpha {\mathrm{\parallel }}_B))$ and [F9] give $\mathrm{\parallel }(1-t)x+ty-p{\mathrm{\parallel }}_B\le (1-t)\mathrm{\parallel }x-p{\mathrm{\parallel }}_B+t\mathrm{\parallel }y-p{\mathrm{\parallel }}_B<1\mathrm{/}(2\mathrm{\parallel }\alpha {\mathrm{\parallel }}_B)$, so $(1-t)x+ty\in U$. List the finitely many walls meeting $U$ as $H_1,\dots ,H_N$. A point of an alcove meeting $U$ determines a sign for each listed wall. If two alcoves meeting $U$ have the same sign vector, take one point from each; every wall outside the list misses $U$, and a segment in the convex set $U$ cannot join opposite sides of a wall without meeting it. The segment between the two points therefore stays in the same strict halfspace for each listed wall and misses every wall outside the list, so it lies in the complement and connects the points. They belong to the same connected component. There are at most $2^N$ sign vectors, hence only finitely many alcoves meet $U$.

The identity, composition and inverses of bijective distance-preserving maps are again bijective and distance-preserving: for a composition this follows by applying the two distance equalities in succession, and for an inverse it follows by writing $x=f^{-1}(u)$ and $y=f^{-1}(v)$ in $d_B(f(x),f(y))=d_B(x,y)$. Thus $\mathrm{I}\mathrm{s}\mathrm{o}\mathrm{m}(E)$ is a group under composition by [F13]. For all $x,y\in E$, $d_B(t_{k{\alpha }^{\vee }}x,t_{k{\alpha }^{\vee }}y)=\mathrm{\parallel }x-y{\mathrm{\parallel }}_B=d_B(x,y)$, and $d_B(s_{\alpha }x,s_{\alpha }y)=\mathrm{\parallel }{s}_{\alpha }(x-y){\mathrm{\parallel }}_B=\mathrm{\parallel }x-y{\mathrm{\parallel }}_B$ because $s_{\alpha }$ is orthogonal by [F1]. By step 2.1, $r_{\alpha ,k}$ is bijective; by step 1.1 it is the composition of those two distance-preserving maps, so it belongs to $\mathrm{I}\mathrm{s}\mathrm{o}\mathrm{m}(E)$. Its linear part fixes $H_{\alpha ,0}$ pointwise and sends $\mathbb{R}\alpha$ to its negative, so its differential acts as $-1$ on the normal line.

For uniqueness, let $g$ be a Euclidean isometry with fixed set $H_{\alpha ,k}$. Put $h=x-\frac{B(x,\alpha )-k}{B(\alpha ,\alpha )}\alpha \in H_{\alpha ,k}$. For every $u$ with $B(u,\alpha )=0$, both $h$ and $h+u$ are fixed by $g$; equality of squared distances from $g(x)$ and $x$ to these two points gives $B(g(x)-h,u)=0$ and $\mathrm{\parallel }g(x)-h{\mathrm{\parallel }}_B=\mathrm{\parallel }x-h{\mathrm{\parallel }}_B$. Write $g(x)-h=c\alpha +u_0$ with $c=B(g(x)-h,\alpha )\mathrm{/}B(\alpha ,\alpha )$ and $u_0\in \{u:B(u,\alpha )=0\}$. Since $\alpha$ is orthogonal to that subspace, $u_0$ is orthogonal to it too; in particular $B(u_0,u_0)=0$, so $u_0=0$. Also write $x-h=t\alpha$. The norm equality gives $c=\pm t$. If $t=0$, then $x=h$ and $g(x)=x$; if $t\mathrm{\ne }0$, the condition $\mathrm{Fix}(g)=H_{\alpha ,k}$ excludes $c=t$, so $c=-t$. Hence $g(x)=h-t\alpha =r_{\alpha ,k}(x)$ for every $x$, proving uniqueness.

If $y\in H_{\beta ,l}$, then orthogonality of $s_{\alpha }$ and $s_{\alpha }({\alpha }^{\vee })=-{\alpha }^{\vee }$ give $B(r_{\alpha ,k}(y),s_{\alpha }\beta )=B(s_{\alpha }y,s_{\alpha }\beta )+kB({\alpha }^{\vee },s_{\alpha }\beta )=l-kB({\alpha }^{\vee },\beta )$. The level on the right is an integer by [F1], and $s_{\alpha }\beta \in \mathrm{\Phi }$ by [F2]; since $r_{\alpha ,k}$ is an involution, this proves the wall-image equality.

Each generator $r_{\alpha ,k}$ therefore permutes the wall arrangement. It is a homeomorphism, so it preserves the complement and maps connected components to connected components; hence $W_a$ permutes the alcoves. For $k=0$, step 1.1 gives $r_{\alpha ,0}=s_{\alpha }$, so the generators of $W$ lie in $W_a$ and permute the walls through the origin.

By step 1.1, $r_{{\alpha }_s,1}\circ r_{{\alpha }_s,0}=t_{{\alpha }_s^{\vee }}$ for each simple root. The simple coroots form a basis of $E$ and generate $Q^{\vee }$ by step 1.5, so the translations $t_{\lambda }$ for all $\lambda \in Q^{\vee }$ lie in $W_a$; also $W\le W_a$ by step 4.1. Therefore every value $\psi (\lambda ,w)=t_{\lambda }\circ w$ lies in $W_a$.

Every generator $r_{\alpha ,k}$ equals $\psi (k{\alpha }^{\vee },s_{\alpha })$ by step 1.1, and each inverse is the same generator by step 2.1. By [F8], every finite word in these images is the image under $\psi$ of the corresponding product in $Q^{\vee }⋊W$. Since such words form $W_a$, this proves $W_a\subseteq \mathrm{im}\psi$. Together with step 5.1 we get $\mathrm{im}\psi =W_a$; injectivity from step 1.4 then gives the claimed unique Euclidean decomposition. All arguments use only finite subcovers, finite lists and finite sums; no choice principle is used.

Let $v$ be the regular vector defining ${\mathrm{\Phi }}^{+}$. Its orthogonal projection to $E_i$ has the same nonzero pairings with roots in ${\mathrm{\Phi }}_i$, so ${\mathrm{\Phi }}_i^{+}={\mathrm{\Phi }}_i\cap {\mathrm{\Phi }}^{+}$ is a positive system on ${\mathrm{\Phi }}_i$. Its simple roots are ${\mathrm{\Delta }}_i=\mathrm{\Delta }\cap {\mathrm{\Phi }}_i$: if $\gamma \in {\mathrm{\Phi }}_i^{+}$ is a sum $\gamma =\beta +\delta$ of positive roots of $\mathrm{\Phi }$, project the equality to each $E_j$ with $j\mathrm{\ne }i$. Each positive root lies in one component; if exactly one summand lies in $E_j$, its projection is nonzero, and if both lie in $E_j$, their sum is nonzero because $B(v,\beta +\delta )>0$. Thus neither summand lies outside $E_i$. Decomposability in ${\mathrm{\Phi }}_i$ is therefore equivalent to decomposability in $\mathrm{\Phi }$. Finally, each root in ${\mathrm{\Phi }}_i$ has only ${\mathrm{\Delta }}_i$ coordinates in the basis $\mathrm{\Delta }$, so ${\mathrm{\Delta }}_i$ spans $E_i$; it is linearly independent as a subset of $\mathrm{\Delta }$. Hence the simple-root basis splits into bases ${\mathrm{\Delta }}_i$ of the $E_i$.

Fix $i$. Write ${\theta }_i={\sum }_{s\in S_i}{n}_{i,s}{\alpha }_s$ with $n_{i,s}\in {\mathbb{Z}}_{\ge 0}$, not all zero, by [F2]. If its support $T=\{s:n_{i,s}>0\}$ were proper, then for every $s\in S_i\setminus T$, dominance from [F3] and nonpositivity [F7] would give $0\le B({\theta }_i,{\alpha }_s)={\sum }_{t\in T}{n}_{i,t}B({\alpha }_t,{\alpha }_s)\le 0.$ Thus every simple root in $T$ is orthogonal to every simple root in $S_i\setminus T$, so by [F8] the connected Dynkin graph would be disconnected. Hence $T=S_i$, and every $n_{i,s}>0$. For $\gamma \in {\mathrm{\Phi }}_i^{+}$, the finite set $P_{\gamma }:=\{\eta \in {\mathrm{\Phi }}_i^{+}:\gamma \le \eta \}$ is nonempty. It has a maximal element $\eta$ because it is finite. If $\eta \le {\eta }^{\mathrm{\prime }}$ for a positive root ${\eta }^{\mathrm{\prime }}$, then transitivity gives $\gamma \le {\eta }^{\mathrm{\prime }}$, so ${\eta }^{\mathrm{\prime }}\in P_{\gamma }$ and maximality forces ${\eta }^{\mathrm{\prime }}=\eta$. Thus $\eta$ is maximal among all positive roots, and uniqueness of the highest root in [F3] gives $\eta ={\theta }_i$. By the root-order definition in [F3], ${\theta }_i-\gamma ={\sum }_{s\in S_i}{m}_s{\alpha }_s$ with $m_s\in {\mathbb{Z}}_{\ge 0}$. For $x\in C$, each $B(x,{\alpha }_s)>0$, so $B(x,\gamma )>0$ by [F2] and $B(x,{\theta }_i)-B(x,\gamma )={\sum }_{s\in S_i}{m}_{s}B(x,{\alpha }_s)\ge 0.$ The dominance inequality $B({\theta }_i,\gamma )\ge 0$ is [F3]. If $B(x,{\theta }_i)<1$, the displayed bound gives $B(x,\gamma )<1$ as well.

Fix $i$, write $n=\mathrm{\mid }{S}_i\mathrm{\mid }\ge 1$, and define the linear map $T_i:E_i\to {\mathbb{R}}^{S_i}$ by $T_i(x{)}_s=n_{i,s}B(x,{\alpha }_s)$. Its kernel is zero: if every coordinate vanishes, then $x$ is orthogonal to the basis ${\mathrm{\Delta }}_i$, hence to all of $E_i$, and positive definiteness gives $x=0$. The domain and codomain both have dimension $n$, so $T_i$ is bijective. Its coordinate functionals are continuous by Cauchy–Schwarz, so $A_i=T_i^{-1}(\{u:u_s>0,{\sum }_s{u}_s<1\})$ is open. It is nonempty, since $u_s=1\mathrm{/}(2n)$ gives a point of it, and it is convex; it is exactly the interior of the closed coordinate simplex below. Its closure is $T_i^{-1}(\{u:u_s\ge 0,{\sum }_s{u}_s\le 1\})$: continuity gives one inclusion; for any $u$ in that closed set, put $x=T_i^{-1}(u)$ and $x^{\ast }=T_i^{-1}(u^{\ast })$, where $u_s^{\ast }=1\mathrm{/}(2n)$. Then $x_t=(1-t)x+t{x}^{\ast }$ lies in $A_i$ for $0<t\le 1$, and $\mathrm{\parallel }{x}_t-x{\mathrm{\parallel }}_B=t\mathrm{\parallel }{x}^{\ast }-x{\mathrm{\parallel }}_B\to 0$, proving the other inclusion. The closed coordinate simplex is the convex hull of $0$ and the standard coordinate vectors $e_s$, which are affinely independent; its inverse image is therefore a geometric simplex. Every point in the closure is ${\sum }_s{u}_s{T}_i^{-1}(e_s)$ with $0\le u_s\le 1$, so by [F9] its norm is at most ${\sum }_s\mathrm{\parallel }{T}_i^{-1}(e_s){\mathrm{\parallel }}_B$. Its barycentric coordinates are $u_s$ and $u_0=1-{\sum }_s{u}_s$. Each of the $n+1$ equalities $u_s=0$ or $u_0=0$ cuts out an $(n-1)$-dimensional face; every boundary point satisfies one of these equalities, so these are exactly the facets. These coordinate facets correspond to $H_{{\alpha }_s,0}$ and $H_{{\theta }_i,1}$.

On $A_i$, step 1.2 gives $0<B(x,\gamma )<1$ for every positive root $\gamma \in {\mathrm{\Phi }}_i^{+}$; for a negative root $-\gamma$ it gives $-1<B(x,-\gamma )<0$. Hence no integer level $H_{\beta ,k}$ for $\beta \in {\mathrm{\Phi }}_i$ meets $A_i$. For nonempty $J$ equal to a singleton $\{i\}$ or to all components $I$, the set $A_J:={\prod }_{j\in J}{A}_j$ is nonempty, convex, and avoids every wall of the arrangement on $E_J:={⨁}_{j\in J}{E}_j$. Let $D_J$ be the connected component of that wall complement containing a point $x\in A_J$. For each simple root ${\alpha }_s$ in these components, the continuous function $y\mapsto B(y,{\alpha }_s)$ has no zero on $D_J$ and is positive at $x$, so its negative and positive preimages cannot both be nonempty, since they would separate $D_J$; it therefore stays positive throughout $D_J$. For each $j\in J$, the function $y\mapsto B(y,{\theta }_j)-1$ likewise has no zero on $D_J$ and is negative at $x$, so it stays negative. Thus $D_J\subseteq A_J$. Since $A_J$ is connected and meets $D_J$, maximality of the connected component gives $A_J\subseteq D_J$, hence $A_J=D_J$. Taking singleton $J$ proves that each $A_i$ is an alcove; taking $J=I$ proves that $A$ is an alcove for the full arrangement when $I$ is nonempty. By [F4] these components are open, as claimed.

If $\mathrm{\Phi }=\mathrm{\varnothing }$, then $E=0$ by [F1], the wall arrangement is empty, and its complement has the single component $\{0\}$. If ${\mathrm{\Phi }}_i=\{\pm {\alpha }_s\}$ is of type $A_1$, its positive roots contain only its simple root ${\alpha }_s$, so ${\theta }_i={\alpha }_s$ and the two inequalities defining $A_i$ are exactly $0<B(x,{\alpha }_s)<1$. The product statement already established applies independently to every component, including any collection of $A_1$ factors. The proof constructs only finite component decompositions, finite coordinate vectors, and finite sums; no axiom of choice is used.